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Chapter 4: Problem 90

A 1.248 -g sample of limestone rock is pulverized and then treated with 30.00 mL of 1.035\(M\) HCl solution. The excess acid then requires 11.56 \(\mathrm{mL}\) of 1.010 \(\mathrm{M}\) NaOH for neutralization. Calculate the percentage by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the HCl solution.

Short Answer

Expert verified
The percentage by mass of calcium carbonate (CaCO3) in the limestone rock sample is approximately 98.4%. This value was obtained by considering the stoichiometry of the involved chemical reactions, calculating the moles of HCl that reacted with CaCO3, determining the moles of CaCO3 in the rock sample, and finally computing the mass and percentage of CaCO3 in the sample.

Step by step solution

01

Write down the balanced chemical equations involved in the reactions

First, we have the reaction between calcium carbonate and hydrochloric acid: CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g) Next, we have the reaction between the excess hydrochloric acid and sodium hydroxide: HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
02

Calculate the moles of HCl reacted with CaCO3 and NaOH

We can calculate the moles of HCl that reacted with the CaCO3 and NaOH. First, let's find the moles of HCl added initially: moles of HCl = volume (L) × concentration (M) moles of HCl = 0.03 L × 1.035 mol/L = 0.03105 mol Next, let's find the moles of HCl that reacted with the NaOH: moles of HCl = volume (L) × concentration (M) moles of HCl = 0.01156 L × 1.010 mol/L = 0.0116736 mol
03

Calculate the moles of HCl that reacted with CaCO3

By subtracting the moles of HCl that reacted with NaOH from the moles of HCl added initially, we can determine the moles of HCl that reacted with the CaCO3: moles of HCl reacted with CaCO3 = total moles of HCl - moles of HCl reacted with NaOH moles of HCl reacted with CaCO3 = 0.03105 mol - 0.0116736 mol ≈ 0.0193764 mol
04

Calculate the moles of CaCO3 in the rock sample

By keeping the stoichiometry of the reaction between CaCO3 and HCl in mind, we can calculate the moles of CaCO3 in the rock sample: CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g) 1 mole of CaCO3 reacts with 2 moles of HCl. Therefore, moles of CaCO3 = moles of HCl reacted with CaCO3 / 2 moles of CaCO3 ≈ 0.0193764 mol / 2 ≈ 0.0096882 mol
05

Calculate the mass of CaCO3 in the rock sample

Now, we can find the mass of CaCO3 in the rock sample: mass of CaCO3 = moles of CaCO3 × molar mass of CaCO3 mass of CaCO3 ≈ 0.0096882 mol × (40.08 + 12.01 + 3 × 16.00) g/mol ≈ 1.228 g
06

Calculate the percentage of CaCO3 in the rock sample

Finally, we can calculate the percentage by mass of CaCO3 in the rock sample: percentage of CaCO3 = (mass of CaCO3 / mass of rock sample) × 100% percentage of CaCO3 ≈ (1.228 g / 1.248 g) × 100% ≈ 98.4% The percentage by mass of calcium carbonate in the rock sample is approximately 98.4%.

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Most popular questions from this chapter

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